题目说给你n个水箱,初始是没有水的,每个的高低位置可能不同,给了你初始的水箱底部所处的水平线位置,问给你V体积水时,水的水平线位置。
直接二分位置p,对于每一个底部低于水平线位置的水箱,里面的水的体积 = min(h,p-b)*w*d;
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 50005 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct node 18 { 19 int b,h,w,d; 20 }p[N]; 21 int dcmp(double x) 22 { 23 if(fabs(x)<eps) return 0; 24 return x<0?-1:1; 25 } 26 double cal(double mid,int n) 27 { 28 int i; 29 double s = 0; 30 for(i = 1; i <=n;i++) 31 { 32 if(p[i].b>mid) continue; 33 s+=min(mid-p[i].b,p[i].h*1.0)*p[i].w*p[i].d; 34 } 35 return s; 36 } 37 int main() 38 { 39 int v,n,i,t; 40 cin>>t; 41 while(t--) 42 { 43 scanf("%d",&n); 44 double s = 0; 45 for(i =1 ; i <=n; i++) 46 { 47 scanf("%d%d%d%d",&p[i].b,&p[i].h,&p[i].w,&p[i].d); 48 s+=p[i].h*p[i].w*p[i].d; 49 } 50 scanf("%d",&v); 51 if(dcmp(v-s)>0) 52 { 53 puts("OVERFLOW"); 54 continue; 55 } 56 double rig = 2000000.0,lef = 0,mid; 57 while(rig-lef>eps) 58 { 59 mid = (rig+lef)/2.0; 60 if(dcmp(cal(mid,n)-v)>=0) 61 rig = mid; 62 else lef = mid; 63 } 64 printf("%.2f\n",rig); 65 } 66 return 0; 67 }
poj1434Fill the Cisterns!(二分)
时间: 2024-10-10 08:25:01